I have a process that will run in the background like a daemon. I want that daemon to give me it's status whenever I send a specific signal to it.
My code for doing so is as follows:

my($Percent) = 50;
$SIG{HUP} = \&status;

sub status () {
print "\n I'm $Percent done.";
}

And I initiate it by sending: kill -s HUP <PID> or in perl: (kill "HUP", <PID>)
However, Although this works, It will kill my process if it's run two times within a short intervall. This is the error I get:

*** sysread(out): Interrupted system call

Q: Why does my process terminate after sending signals two times?
Q: Which signal is the best to use for something like printing status-info?
Q: Is it possible to make a new signal like "SIGSTATUS" ?

In reply to How to create private signals? (or block interruption) by rapide

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